Question

Find the equation of a line passing through the point (5,-1) and perpendicular to the line 2x-3y-4=0

Answer 1

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As it is perpendicular to 3x+4y-5=0, the product of slopes of both the lines should be -1.

Let m1 be the slope of the given line and m2 slope of the equation of the line to be found. Slope of the given line is -3/4. So, -3/4×m2=-1, m2= 4/3.

Now we know the slope of the line and one point is given, by applying the formula, y-y1=m(x-x1)

y−3=4/3(x−2)y−3=4/3(x−2)

3y−9=4x−83y−9=4x−8

4x−3y+1=0.4x−3y+1=0.

It is the required equation :)

Answer 2

$\textbf {Hey there, here is the solution.}$

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2x - 3y - 4 = 0 (Given)

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Find the gradient of the perpendicular line:

2x - 3y - 4 = 0

3y = 2x - 4

y = 2/3 x - 4/3

⇒ gradient = 2/3

⇒ gradient of the perpendicular line = -3/2

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Find the y-intercept of the line:

It passes through (5, - 1) {given)

y = mx + c

-1 = -3/2 (5) + c

-1 = -15/2 + c

c = 13/2

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Equation: y = -3/2 x + 13/2

Answer: The equation is y = -3/2 x + 13/2

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⭐$\textbf {Cheers}$⭐